```Question 109478
Your equation is incorrect. Assuming that the diver is in free fall (meaning that he
had no initial velocity) the equation should be:
.
h = -4.9t^2 + 21
.
In which h is the height of the diver in meters and t is the time the diver has fallen. Note
that at time zero, you can set t equal to zero and the equation reduces to h = 21. This
means that just before he jumps, the diver is 21 meters above the water ... just as the
problem tells you he should be.
.
How long does it take to reach the water?  When the diver just reaches the water his height
is zero. So substitute zero for h and you have:
.
0 = -4.9t^2 + 21
.
Subtract 21 from both sides to get rid of the +21 on the right side and you have:
.
-21 = -4.9t^2
.
now divide both sides by -4.9 and the result is:
.
t^2 = -21/-4.9 = 4.2857
.
Finally, solve for t by taking the square root of both sides to get:
.
t = sqrt(4.28570 = 2.0702 seconds
.
So the diver will hit the water 2.0702 seconds after leaving the platform.
.
When the diver is halfway down, his height will be half of the 21 meters. So the diver's height
will be 10.5 meters. Substitute this into the height equation and you have:
.
10.5 = -4.9t^2 + 21
.
Subtract 21 from both sides to get rid of the 21 on the right side and you have:
.
-10.5 = -4.9t^2
.
Divide both sides by -4.9 and the result is:
.
t^2 = -10.5/-4.9 = 2.1429 seconds
.
Find t by taking the square root of both sides and you have:
.
t = 1.4639 seconds
.
So in 1.4639 seconds the diver is half way down. And in the remaining time of 0.6063 seconds
(comes from 2.0702 seconds to fall the entire 21 meters less the 1.4639 seconds to fall
the first half of the way) the diver falls the rest of the way. His rate of fall is getting
faster and faster the more time that he is falling.
.
Hope this helps you to see your way through the problem.
.
```